Singularity proofs

dirty l3um · **Joined:** Wed, 9 Jan 2008 19:24:20 UTC **Posts:** 8

Alright im having some issues here with proofs (i think). I have the following problem.

Code:

Let A and B be nxn matrices and let C = A-B. Show that if A(Xo) = B(Xo) and (Xo) !=0 then C must be singular

I have attempted to prove it and am unsure if i have come to the correct conclusion, or if im even mathematically allowed to do what i have done

C = A - B
A(Xo) = B(Xo)
A(Xo)-B(Xo) = 0

C(Xo) = (A-B)Xo
C(Xo) = 0

Since Xo cannot be zero than C must be singular. Does that make sense???

In addition to that problem I have another one that im not sure if the proof is correct. The problem states....

Code:

Let A and B be nxn matrices and let C=AB. Prove that is B is singular then C must also be singular.

Attempt at a proof....

C=AB
I=I
C(C^-1) = I
AB(C^-1) = I
C^-1 = I(AB)^-1
C^-1 = B^-1 * A^-1

Since we know that B is singular B^-1 is undetermined, meaning that C^-1 must also be undetermined.  And that proves that C must be singular????  I think?

I feel like neither answer is correct....

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

dirty l3um wrote:

Let A and B be nxn matrices and let C = A-B. Show that if A(Xo) = B(Xo) and (Xo) !=0 then C must be singular

C = A - B
A(Xo) = B(Xo)
A(Xo)-B(Xo) = 0

C(Xo) = (A-B)Xo
C(Xo) = 0

In order for C to be nonsingular, its kernel must be trivial (i.e., contain only the zero vector).

dirty l3um wrote:

Let A and B be nxn matrices and let C=AB. Prove that is B is singular then C must also be singular.

Because B is singular, there is a nonzero vector x in its kernel, so Bx = 0. Consequently Cx = ABx = 0, so x belongs to the kernel of C also.

dirty l3um · **Joined:** Wed, 9 Jan 2008 19:24:20 UTC **Posts:** 8

so does that mean that if in the first case to prove it is singular i could do something like.....
(X1 being a non zero number...)

A(Xo) = B(Xo)
(A-B)(Xo) = 0

C = A - B
C(Xo) = X1
(A-B)(Xo) = X1

this is a contradiction???? If not why not? Im very confused at this point.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

It's a contradiction because, for any nonsingular matrix C, we have C(Xo) nonzero if and only if Xo is nonzero.

dirty l3um · **Joined:** Wed, 9 Jan 2008 19:24:20 UTC **Posts:** 8

Matt, i think i understand the basis of what you are saying, however im having a very difficult time transposing that into a mathematical proof.... How would i prove it? (with equations)

Shadow · **Posted:** Thu, 27 Jan 2011 19:31:12 UTC

dirty l3um wrote:

Matt, i think i understand the basis of what you are saying, however im having a very difficult time transposing that into a mathematical proof.... How would i prove it? (with equations)

What do you mean "[transcribing] that into a mathematical proof?" that is a proof. It's nothing but definitions, so it should be transparent.

By definition, if B is singular, there is $x_0\ne 0$ such that Bx_0=0

. But then Cx_0=ABx_0=A(Bx_0)=A(0)=0

, so C is also singular, with a singular vector x_0

dirty l3um · **Joined:** Wed, 9 Jan 2008 19:24:20 UTC **Posts:** 8

Thankyou for your help shadow and matt. I believe the problem was that i wasnt entirely sure what a proof was

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